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Unformatted text preview: CHAPTER 9 LARGE—SAMPLE TESTS
OF HYPOTHESES 9.1 Testing Hypotheses about Population Parameters Often statisticians are not interested solely in estimating unknown population parameters,
but in deriving the formulation of rules or procedures that leads to a decision culminating in
the acceptance or rejection of some statement or hypothesis about the population parameters.
In light of sample observations, sometimes the problem is to decide which of several assertions
or hypotheses is most supported by the sample evidence For example, 1.. an educator might be interested in deciding on the basis of sample data whether or
not male students are in general doing better than female students in mathematics; 2. a medical researcher might have to decide on the basis of sample evidence whether or
not a new drug reduces the blood pressure to normal levels; 3. an environment advocator might wish to collect appropriate sample data to decide
whether or not the level of benzene in a speciﬁc oil reﬁnery is systematically above the com
pliance standard set by government for benzene levels. The area of statistical test of hypothesis comprises a major area of statistical inference in
developing rules or procedures that lead to the decision on accepting or rejecting assertions
or hypotheses such as those presented in the above example. 9.2 A Statistical Test of Hypothesis A statistical test of hypothesis consists of ﬁve parts:
1. The null hypothesis, denoted by H0. The alternative hypothesis, denoted by H a. The test statistic and its p~value The rejection region. The conclusion.
First, let us deﬁne precisely the terms of statistical hypothesis and hypothesis testing. Definition 1,. A statistical hypothesis is an assertion or conjecture concerning one or
more populations, speciﬁcally about one or more parameters of such population distribu—
tions. The hypothesis testing is a statistical procedure that is designed to test a statistical
hypothesis. Deﬁnition 2. The two competing hypotheses are the alternative hypothesis Ha, gen—
erally the hypothesis that the researcher wishes to support, and the null hypothesis H0, a
contradiction to the alternative hypothesis Ha. Usually, the statistical test begins by assuming that the null hypothesis H0 is true. Then
the researcher uses the sample data to decide whether the evidence favors Ha rather than
H0 and draws one of these two conclusions: * Reject H0 and conclude that Ha is true.
* Accept (do not reject) H0 as true. Example 1. We wish to show that the average hourly wage of construction workers
in the state of California is different from $14, which is the national average. This is the alternative hypothesis, written as
Ha : ,u 7E 14 The null hypothesis is
H0 2 ,U, = We would like to reject the null hypothesis, thus concluding that the California mean is not
equal to $14. Example 2. A milling process currently produces an average of 3% defectives, We are
interested in showing that a simple adjustment on a machine will decrease p, the proportion
of defectives produced in the milling process. Thus, the alternative hypothesis is Ha : p < 0.03 and the null hypothesis is
H0 : p = 0.03 If we can reject the null hypothesis HO, we can conclude that process produces fewer than
3% defectives. Example I gives a two—tailed test of hypothesis while Example 2 gives a one—tailed test of
hypothesis. The decision to reject or accept the null hypothesis is based on information con—
tained in a sample drawn from the population of interest. This information takes these forms: * Test statistic: a single number calculated horn the sample data. * pv’ualue: a probability calculated using the test statistic. Example 3. For the test of hypothesis in Example 1, the average hourly wage E for a
random sample of n = 100 California construction workers might be a good test statistic for testing the null hypothesis is
H0 I M = versus the alternative hypothesis
Ha : ,a 71$ 14,. If the null hypothesis H0 is true, then the sample mean should not be too far from the
population mean a = 14.. Suppose that this sample produces a sample mean 5 = 15 with
standard deviation 3 = 2.. Is this sample evidence likely or unlikely to occur, if in fact H0
is true? We can use two measures to ﬁnd out. Since the sample size is large, the sampling
distribution of E is approximately normal with mean a = 14 and standard error a/ﬁ,
estimated as
' s 2
SE~~=—=O.2. \/ﬁ «100
* The test statistics 5 = 15 lies
IE — a 15 —— 14 3N5 :Efﬁb‘zo 2 standard deviations from the population mean a. * The p~value is the probability of observing a test statistic as extreme as or more
extreme than the observe value, if in fact H0 is true. For this example, we define “extreme” as far below or far above what we would have expected. That is, 5. 2% p—value=P{z > 5} + P{z <: —5} m 0. The large value of the test statistic and the small p—value mean that we have observe a
very unlikely event, if indeed H0 is true and a = 14. Deﬁnition 3. The set, consisting of values that support the alternative hypothesis and
lead to rejecting H0, is called rejection region. The other, consisting of values that support the null hypothesis, is called acceptance region. In Example 1, we would be inclined to believe that the average hourly wage of construction
workers in California is different from $14 if the sample mean is either much less than $14
or much greater than $14., As a result, a two—tailed rejection region as shown below will be appropriate. Rejection ; Acceptance ; Rejccﬁon
region ; region : region Critical value —j $14 Critical value In Example 2, the intention is to claim that the percentage of defectives has decreased so
that we would be inclined to reject the null hypothesis for values of the sample proportion
much smaller than 0.03. Thus, a left—tailed rejection region as shown below will be proposed. Acceptance Rejection
region region ' __l
. . J .03
CIlUCal value "tb Deﬁnition 4. The test is called a twotailed test if the rejection region is in both tails
of the distribution. When the rejection region is in the left tail of the distribution, the test is
called a left—tailed test. A test with its rejection region is in the right tail of the distribution
is called a right—tailed test. If the test statistic falls into the rejection region, then the null hypothesis is rejected. If
the test statistic falls into the acceptance region, then either the null hypothesis is accepted
or the test is judged to be inconclusive. How do we decide on the critical value that separate the acceptance and rejection regions?
This depends on the amount of conﬁdence that we want to attach to the test conclusions
and the signiﬁcance level, the risk we are willing to take of making in incorrect decision. Deﬁnition 5. A Type I error for a statistical test is the error of rejecting the null
hypothesis when it is true. The level of signiﬁcance (or signiﬁcance level) for a statistical
test of hypothesis is or = P {Committing Type I error}
: P {Falser rejecting H0}
P {Rejecting H0 when it is true} . This value (1 represents the maximum tolerable risk of incorrectly rejecting H0, Once
this signiﬁcance level is ﬁxed, the rejection region can be set to allow the researcher to reject
Hg with a ﬁxed degree of conﬁdence in the decision. 9.3 A LargeSample Test about A Population Mean Consider a random sample of n measurements drawn from a population that has mean
,a and standard deviation 0,. We want to test the null hypothesis H0 3 M 2 #0
where no is some hypothesized value for a, versus a onetailed alternative hypothesis
Ha : a > no Note that H0 provides an exact value for the parameter to be tested, whereas Ha gives a
range of possible values for a Naturally, the sample mean ‘55 is the best estimate of the actual value for a The values
of E that are extremely large would imply that is greater than n hypothesized. Hence, we
should reject H0 if 3c" is too large. Values of E that lie too many standard deviations to the right of the mean are not likely
to occur. Those values have very little area to their right. Hence, we deﬁne “too large” as being too many standard deviations away from not. Since the sampling distribution of the sample mean E is approximately normal when n
is large, the number of standard deviations that E lies from no can be measured using the standardized test statistic _
53 * M0 s/Jﬁ ’
where s/ is the estimated standard error of E, which has an approximately standard
normal distribution when H0 is true and ,u = no. The significance level a is equal to the area under the normal curve lying above the rejection region. Thus, if we want to a = 0.01, we
will reject Ho when if is more than 2.33 standard deviations to the right of no. Equivalently, we will reject H0 if the standardized test statistic z is greater than 2.33. 2% f(z) 0 233 Z
Accaptance region ——————>'¢ Rejection region Example 4. The average weekly earning for women in managerial and professional
positions is $670. Do men in the same positions have average weekly earnings that are higher
than those for women? A random sample of n = 40 men in managerial and professional
positions showed E = $725 and s = $102.. Test the appropriate hypothesis using a = 0.01. Solution.
1. We want to test the null hypothesis H0:n=670 2. A onetailed alternative hypothesis is Hazn>670 3. The test statistic is
_'E—,u0 _ 725 —670 Z s/x/ﬁ _ 102N216 4. Rejection Region: For CM : 0.01, we will reject H0 if the standardized test statistic z
is greater than 2.33. = 3.41. 5. Since 341 > 2.33, we reject H0 and conclude that the average weekly earnings for
men in managerial and professional positions are higher than the average for women. The
probability that we have made an incorrect decision is a = 0.01.. If we want to test the null hypothesis
Ho : M = #0 where ,uo is some hypothesized value for 11, versus a two—tailed alternative hypothesis
Ha 3 I” 75 M0 The alternative hypothesis implies either a > no or ,u < #0. Values of E that are either “too
large” or “too small” in terms of their distance from no are placed in the rejection region. For oz : 0.01, we will reject H0 if the standardized test statistic z is greater than 2575 or
smaller than —2H575. 01/2 = 005 —2 ‘5 8 0 2 58 3
Rejection region—>1 l<——Rejection region Example 5. The daily yield for a local chemical plant has average 880 tons for the
last several years. The quality control manager would like to know whether this average
has changed in recent months. She randomly selects 50 days from the computer database
and computes the average and standard deviation of the n = 50 yields as E = 871 tons and
s = 21 tons, respectively. Test the appropriate hypothesis using a = 0105. Solution.
1. We want to test the null hypothesis H0 : n = 880
2. A twotailed alternative hypothesis is
Ha : M # 880 3. The test statistic is
E — no 871 — 880
z z ——— = —————— = +3.03.
s/ﬁ 21 N50
4. Rejection Region: For a = 01,05, we will reject H0 if the standardized test statistic z
is either greater than 1.96 or smaller than —1.96. 5‘. Since ——3.03 < —1.,96, we reject HO and conclude that the average of daily chemical
yield has changed. The probability that we have made an incorrect decision is a = 0.05. Large—Sample Statistical Test for a
l. Null hypothesis: H0 : a = a0
2. Alternative hypothesis:
Ha : ,u >' no (Upper—Tailed Test) Ha : a < #0 (Lower—Tailed Test)
Ha : a 75 p0 (Two—Tailed Testag) 3. Test Statistic: _ _ E __
z = (fr/J20 estimated as 2 a: 8 / «1:70
4., Rejection Region:
z > 20, for Ha : ,u > no (Upper—Tailed Test)
z < —za for Ha : u < ,uo (Lower—Tailed Test) z > Za/g or z < —za/2 for Ha : ,u # M0 (Two—Tailed Testz) Assumptions: The sample of size n is selected at random from the population and n is large — say, n 2 30,. Calculating the p—‘Value In the previous examples, the decision to reject H 0 was made by comparing the calculated
value of the test statistic with a critical value of 2 based on the signiﬁcance level a of the
test. Different signiﬁcance levels may lead to different conclusions. For example, if in a
righttailed test, the test statistic z = 2.03, we can reject H0 at the 5% level of signiﬁcance
because the test statistic is less than 2 = 233 (see the ﬁgure below). To avoid any ambiguity
in their conclusions, some experimenters prefer to use a variable level of signiﬁcance called the p—value for the test,
f(z) 0 1,645 2,03 2.33 Z Deﬁnition 6. The p—value or observed signiﬁcance level of a statistical test is the small~
est value of a for which H0 can be rejected. It is the actual risk of committing a Type I
error, if H0 is rejected based on the observed value of the test statistic. Note. The p—value measures the strength of the evidence against H0. Example 6. In the right—tailed test for H0 : ,u = no versus H, : [1, > MD with observed
test statistic z 2 2.03, the smallest critical value we can use and still reject H0 is z = 2.03. For this critical value, the risk of an incorrect decision is
P{z '2 2.03}=1— P{z < 2.03} = 1 — 0.9788 2 0.0212. This probability is the p—value for the test. Note that it is actually the area to the right of
the calculated value of the test statistic. Note. A small p—value indicates that the observed value of the test statistic lies far away
from the hypothesized value of a. This presents strong evidence that is false and H0 should
be rejected. Larger p—values indicates that the observed value of the test statistic is not far from the hypothesized mean and does not support rejection of H0. How small does the
p—value need to be before H0 can be rejected? Deﬁnition 7. If the rigvalue is less than or equal to a preassigned signiﬁcance level a,
then the null hypothesis H0 can be rejected, and we can report that the results are statisti— cally signiﬁcant at level a. Example 7. Refer to Example 5. The quality control manager wants to know whether
the daily yield at a local chemical plant — which has average 880 tons for the last several
years — has changed in recent months. A random sample of 50 days gives an average yield of
871 tons with a standard deviation of 21 tons. Calculate the p—value for this two—tailed test of hypothesis. Use the p—value to draw conclusions regarding the statistical test. Solution. The rejection region for this two—tailed test of hypothesis is found in both
tails of the normal probability distribution. Since the observed value of the test statistic is
:r = —3.03, the smallest rejection region that we can use and still reject H0 is [2! > 3.03. For
this rejection region, the value of or is the p—value: p—value : P{z > 3.03} + P{2 < ——3.03}
= 0.0012 + 0.0012 2 00024 Note that the two—tailed p—value is actually twice the tail area corresponding to the calculated
value of the test statistic. If this p—value = 00024 is less than or equal to the preassigned
level of signiﬁcance oz, H0 can be rejected. For this test, we can H0 reject at either the 1% or the 5% level of signiﬁcance] Use of pvalue * If the pvalue is less than 0.01, H0 is rejected. The results are highly signiﬁcant. * If the p—Value is between 0.01 and 005, H0 is rejected. The results are statistically
signiﬁcant. * If the p—value is between 0.05 and 010, H0 is usually not rejected. The results are
only tending toward statistical signiﬁcance. * If the p—value is greater than 0.10, H0 is not rejected. The results are not statistically
signiﬁcant. Example 8 .. Standards set by government agencies indicate that Americans should not
exceed an average daily sodium of 3300 milligrams (mg) .. To ﬁnd out whether Americans
are exceeding this limit, a sample of 100 Americans is selected, and the mean and standard
deviation of daily sodium intake are found to be 3400 mg and 1100 mg, respectively. Use a = 0.05 to conduct a test of hypothesis.
Solution. The hypotheses to be tested are H0 : ,a = 3300 vs. Ha : a > 3300 and the test statistic is __ . £— MO} _ 3400 — 3300 Z N 5N7? 1100/«100 Use oz = 0.05, za 2 1.645. Since 0.91 < 1.645, we do not reject at signiﬁcance level of
a = 0.05.. The two approaches developed in this section yields the same conclusions. * The critical value approach: Use a = 0.05, 2.1 = 1645. Since 0.91 < 1.645, we do not
reject H 0 at signiﬁcance level of a = 0.05. = 0.91. .ﬂZJ p—value = .1814 0 .91 1.645 ' z
L—————>
Reject H0 (z > 1 .645) * The p—value approach: The p—value in this case is the probability that :v is greater
than 0.91:
p—value : P{Z > 0.91} = 1 — 0.8186 = 0.1814. The null hypothesis HO can be rejected only if the p—value is less than or equal to the speci
ﬁed 5% signiﬁcance level, Therefore, H0 is not rejected and the results are not statistically
signiﬁcant (see the ﬁgure above). There is not enough evidence to indicate that the average daily sodium intake exceeds 3300 mg.. TWO Types of Errors Since there are two choices in a statistical test, there are also two types of errors that
can happen as shown in the table below. Null Hypothesis _ Decision True False
Reject H0 Type I error Correct decision
Accept H0 Correct decision Type II error Deﬁnition 8. A Type I error for a statistical test is the error of rejecting the null hypoth
esis Ho when it is true. The probability of making a Type I error is denoted by the symbol a. Deﬁnition 9. A Type II error for a statistical test is the error of accepting the null
hypothesis H0 when it is false and some alternative hypothesis H, is true. The probability
of making a Type II error is denoted by the symbol B. Notice that the probability of a Type I error is exactly the same as the level of signiﬁcance
or and is therefore controlled by the researcher. The Power of a Statistical Test The goodness of a statistical test is measured by the size of the two error rates: a, the
probability of rejecting H0 when it is true, and B, the probability of accepting H0 when H0
is false and Ha is true. A "good" test is one for which both of these error rates are small.
The experimenter begins by selecting oz, the probability of a Type I error. If he decides to
control the value of B, the probability of accepting H0 when H, is true, then an appropriate
sample size is chosen. Another way of evaluating a test is to look at the compliment of Type II error — that is,
rejecting Ha when H, is true — which has probability 1 — ﬂ = P {Rejecting H0 when Ha is true} .. The symbol 1— ﬂ is called the power of the test because it measures the probability of taking
the action that we wish to have occur — that is, rejecting the null hypothesis H0 when it is false and H, is true. Deﬁnition 10.. The power of a statistical test, given as
1 —— B = P {Rejecting H0 when Ha is true} . measures the ability of the test to perform as required. A graph of 1 — B, the probability of rejecting the H0 when in fact H0 is false, as a function
of the true value of the parameter of interest is called the power curve for the statistical test.
Ideally, we would like or to be small and the power 1 — B to be small. Example 9.. Refer to Example 5. Calculate B and the power of the test 1 — B when p
is actually equal to 870 tons. 10 Solution. The acceptance region for the test of Emample 5 is located in the interval 537:1.96i ﬂ Substituting numerical values, we get 21
880 i 1.96——
V50 or
874.18 to 885.82 The probability of accepting , given ,u = 870, is equal to the area under the sampling
distribution for the test statistic E in the interval from 874.18 to 885.82. Since E is normally distributed with ,u = 870 and SE = 21 / V56 2 2.97, 3 is equal to the area under the normal
curve with ,u = 870 located between 874.18 and 885.82 (see the ﬁgure below). Calculating the z—values corresponding to 874.18 and 885.82, we get
_ E —— M 874.18 — 870 = 1.41 z _ = ———
1 5/75 Zl/V/5_0
and _‘ 885 82 870
:c —r ,u . —
z = ~———— = ——r = 5.33.
2 s/ﬁ 21 N50
for)
87 87418 u0=880 885,82 ' f
R8823“ ' Acf§§f§§°e—>l<—Rii:i§§”
Then B = {Accepting H0 when u = 870} P {874.18 < r < 885.82 when it = 870}
P{1.41 < z < 5.33} P {z < 5.33} — P {z < 1.41} 1 — 0.9207 0.0793. Hence, the power of the test is 1 — 8 = 1 8— 0.70793 = 0.9207. 11 The probability of correctly rejecting H0, given that u is actually equal to 870, is 09207”. Values of 1 — [3 can be calculated for various values of ya different from Mo = 880 to
measure the power of the test, For instance, in Ezrample 9, if ,ua = 885, E = {Accepting H0 when u = 870}
= P {87418 < E < 88582 when ,u = 885}
= P{—3n64 < z < 028}
= P{z < 028} — P{2 < #364}
= 0.6103 —— 0
= 0.6103. and the power 1 — ﬂ = 0.3897. The table below shows the power of the test for various
values of ha, and a power curve is graphed in the ﬁgure below. Note that the power of the
test increases as the distance between pa and MO increases. The result is a U —shaped curve for this two tailed test. 0.7673
0,9207
0,9990 Power, 1 — ,5
L0 \ ‘8 2E
'L.._J I'LIlI_L 865 870 875 880 885 890 895 l1 There are many important links among the two error rates, a and ﬂ, the power 1 — ﬂ,
and the sample size, n. Look at the two curves shown in the ﬁgure on page 11. * If or (the sum of the two tail areas in the curve on the right) is increased, the shaded
area corresponding to 5 decreased, and vice versa. * The only way to decrease B for a ﬁxed a is to “buy” more information — that is,
increase the sample size n. What would happen to the area B as the curve on the left is moved closer to the curve
on the right (to = 880)? With the rejection region in the right curve ﬁxed, the value of B will 12 increase. What effect does this have on the power of the test? Look at the power curve in
the above ﬁgure. We may also want to use the Power of a z—Test applet to help us visualize the following
statements: * As the distance between the true (Ma) and the hypothesized (#0) values of the mean
increases, the power 1 — ﬂ increases. the test is better at detecting differences when the distance is large. * The closer the true value (,ua) gets yo the hypothesized value (no), the less power
1 — B the test has to detect the difference. * The only way to increase the power 1 — B for a fixed a is to “buy” more information
— that is, increase the sample size n. The experimenter must decide in the values of oz and B — measuring the risks of the
possible errors he can tolerate. He also must decide how much power is needed to detect
differences that are practically important in the experiment. Once these decisions are made,
the sample size can be chosen by consulting the power curve corresponding to various sample sizes for the chosen test. HOMEWORK: 1919.360 —— 362 9.1, 9.3, 9.9, 9.11, 9.13, 9.15, 9.17 9.4 A Large—Sample Test of Hypothesis for the Difference Between Two Pop—
ulation Means The statistic that summarizes the sample information regarding the difference in popula—
tion means ,ul — M2 is the difference in sample means El — E2. Therefore, in testing whether
the difference in sample means indicates that the true difference in population means differs
from a speciﬁed value, #1 —— #2 = Do, we can use the standard error of f1 — E2 : 2 2 2 2
0 0 _ s s
“l + —2 estimated by —1 + “3
n1 712 n1 H2 in the form of a z—statistic to measure how many standard deviations the difference 51 — E2
lies from the hypothesized difference D0. Large—Sample Statistical Test for M1 — ,LL2 1. Null hypothesis: H0 : M1 — M = D0 13 2. Alternative hypothesis: Ha : a1 — M > DO (Upper—Tailed Test)
Ha : a1 — a2 < D0 (Lower~Tailed Test)
Ha : al —v a2 # D0 (Two—Tailed Test) 3. Test Statistic: 1— _— _.D ——— —D
z = m—ij—g estimated as z % (ac—l—gSELTQ
a+a Va+a
4. Rejection Region:
2 > 2a for Ha : a1 — a2 > Do (Upper—Tailed Test)
2 < —za for Ha : a1 — [12 < D0 (Lower—Tailed Test) z > za/g or z < —za/2 for Ha : ,ul — a2 # D0 (TwoTailed Testz) Assumptions: The samples of size 121 and mg, respectively, are selected randomly and
independently from the two populations and n1 2 30 and n2 2 30.. Example 10. To determine whether car ownership affects a student’s academic achieve—
ment, two random samples of 100 male students were each drawn from the student body.
The grade point average for m = 100 non—owners of cars had an average and variance equal
to E1 = 2.70 and 3% : 0.36, while '52 = 2.54 and 33 = 0.40 for n2 = 100 car owners. Do the
data present sufﬁcient evidence to indicate a different in the mean achievement between car
owners and non—owners of cars? Test using a = 0.05. Solution. Let the mean achievements of car owners and non—owners of car be denoted
by M and a2 , respectively. Then we need to test H02a1—a2=0 vs Hazal—a27é0. The test statistic is
(051 — "252) — D0 _ _2.70 — 2.54 2 R5 2 2 = 1.84..
i 52' 0.36 0.40
m + m 100 + 100 * The critical value approach: Since a = 0.05, 2042 = 1.96. We reject H0 if z > 1.96 or
z < ——1.96. Since —.196 < 1.84 < 196, HO cannot be rejected (see the ﬁgure below). There
is no sufﬁcient evidence to declare a difference in the average academic achievements for the two groups. 14 * The pvalue approach: The pvalue in this case is the probability that z is greater than
2 = 184 plus the probability that z is less than 2 = ——.184, as shown in the above ﬁgure.
Since the p—value of 0.0658 exceeds the speciﬁed signiﬁcance level a = 005, HO cannot be rejected.
p—value = P{z > 1.84} + P{z < —1.84} = (1— Pa $1.84})+ P{z < 4.84}
(1 — 0.9671) + 00329 = 0.0658. ll f(z) :—1 84 0 1 84: 2
§___' ___)
Reject Ho (2 < —1.96) Reject H0 (2 > 1 96) Hypothesis Testing and Conﬁdence interval * If the conﬁdence interval contains the value of the parameter speciﬁed by H0, then
that value is one of the likely or possible values of the parameter and H0 should not be rejected.
* If the hypothesized value lies outside of the conﬁdence limits, H0 is rejected at the a
level of signiﬁcance. Example 11. Refer to Example 10. Construct a 95% conﬁdence interval for the dif—
ference in average academic achievements between car owners and non—owners. Using the
conﬁdence interval, can we conclude that there is a difference in the population means for the two groups of students?
Solution. The point estimate of M1 —— M2 is El — T2 = 2.70 — 2.54 = 0.16 and the standard error is 2 2
81 32 0.36 0.40
,/_ _ = 1/._._. — = 0.0872.
m + mg 100 + 100 Since 1 —— a = 0.95 or or = 0.05, we have Za/g = 1.96. Thus, the 95% conﬁdence interval is 3i 33
_ —_ :l: 1.96 — —
(331 502) ( ) m + m 15 0.16 :E (10.96) (0.0872)
016 i: 0.17 or —0,,01 < ,u1 —— #2 < 0.33. This interval gives us a range of possible values of the difference
in the population means. Since the hypothesized difference, ,ul — a2 = O, is contained in the
conﬁdence interval, we should not reject H0. HOMEWORK: 19190366 — 368 9.19, 9.21, 9,23, 925, 9.27, 929 9.5 A LargeSample Test of Hypothesis for a Binomial Proportion When a random sample of n identical trials is drawn from a binomial population, the
sample proportion if has an approximately normal distribution when n is large, with mean p and standard error __
SE = m
n To test a null hypothesis H0 of the form
H0 3 I? = P0
versus a one» or two~tailed alternative
Ha :p>po or Hatp<po or Ha #979290 the test statistic is constructed using 13, the best estimator of the true population proportion
p. That is, the test statistic is LargeSample Statistical Test for p
1. Null hypothesis: H0 : p = pg
2. Alternative hypothesis:
H, : p > p0 (Upper—Tailed Test) Ha : p < p0 (Lower—Tailed Test)
Ha : pyépo (Two—Tailed Test:) 3. Test Statistic: A
estimated as ,2: = p — po 0
/ 3 P040
’I’L TL 16 4. Rejection Region: 2 > 2a 7 for H, : p > p0 (UpperTailed Test)
2 < —za for Ha : p < p0 (LowerTailed Test)
2 > Za/g or z < ~20,” for Ha : p # p0 (Two—Tailed Test:) Assumptions: The sampling satisﬁes the assumptions of a binomial experiment, and
n is large enough so that the sampling distribution of p can be approximated by a normal distribution (that is, npO > 5 and nqo i> 5). Example 1.2. Regardless of age, about 20% of American adults participate in ﬁtness
activities at least twice a week. However, these ﬁtness activities change as the people get
older, and occasionally participants become nonparticipants as they age. In a local survey
of n = 100 adults over 40 years old, a total of 15 people indicated that they participated in
a ﬁtness activity at least twice a week. Do these data indicate that the participation rate for
adults over 40 years of age is signiﬁcantly less than the 20% ﬁgure? Calculate the p—value and use it to draw the appropriate conclusions.
Solution. We need to test the null hypothesis H0 :p=0.2 versus a one~tailed alternative
Ha : p < 0.2. The test statistic is A
_ p —p0 0.15 — 0.20 1ﬂak : (O.20)(0.80)V
V 100 * The critical value approach: If we choose a = 0.05, then 20, = 1.645. We reject H0 if
z < —1.645. Since ~1.25 > 1.645, H0 cannot be rejected. There is no sufﬁcient evidence to declare that the participation rate for adults over 40 years of age is signiﬁcantly less than
the 20% ﬁgure. 2 —1.25. * The pvalue approach: The p—value is
p—Value = P{z < —1.25} = 0.1056. Since pvalue= 01056 > 0.10, we would not reject H0. There is insufﬁcient evidence to
conclude that the percentage of adults over age 40 who participate in ﬁtness activities twice
a week is less than 20%. 17 f(z) p—value = .1056 HOMEWORK: pp.371 — 373
9.31, 9,33, 9,35, 9,37, 9.39, 9.41 9.6 A Large—Sample Test of Hypothesis for the Difference Between TWO Bi—
nomial Proportions When random and independent samples are selected from two binomial populations, the
difference in the sample proportions 3’51 — 132 has an approximately normal distribution when
sample sizes are large, with mean p1 — p2 and standard error n1 712
To test a null hypothesis H0 of the form
H03p1=P2 0r Hoipi—P2=0
versus a one— or twotailed alternative
Ha1P1>P2 0T Ha3P1<P2 0r Ha1p17ép2 the test statistic is constructed using the pooled estimate $1+$2
n1+n2 13‘: because when H0 is true, the two population proportions are equal to some common value  say, 1)., The test statistic is
p1 ‘ P2 Mal1%) Large—Sample Statistical Test for p1 — p2 18 l. Null hypothesis: H0 :p1 —p2 = 0 or H0 :p1 = pg
2. Alternative hypothesis: Ha : p1 — p2 > 0 (UpperTailed Test)
Ha : p1 — p2 < 0 (Lower—Tailed Test)
Ha : p1 — p2 7A 0 (Two—Tailed Test:) 3.. Test Statistic: _ _ 231 —P2 _ 2/51 —f7\2 _ 171 m P2
an M2 2‘1, ‘ 21
m + M n1+n2 pq(;}1+n—12)
where
A 1171 A :32
P1 = "—" and P2 = —v
711 712 Since the common value of p1 = 192 = p is unknown, it is estimate by ﬁ_ £171 + $2
711 + 712
and the test statistic is
Z = $292— estimated as Z w pats—me) m(%'+%)
4. Rejection Region:
2 > 2a for H, : p1 — p2 > 0 (Upper—Tailed Test)
2 < —2a for Ha : p1 — p2 < 0 (Lower—Tailed Test) 2 > 2042 or z < —za/2 for H, : p1 —p2 71$ 0 (Two—Tailed Testz) Assumptions: Sampling are selected randomly and independently from two binomial
populations, and m and 712 are large enough so that the sampling distribution of 131 — 132
can be approximated by a normal distribution. That is, nlpl > 5, nlql > 5, mm > 5, and TLqu > Example 13. The records of a hospital show that 52 men in a sample of 1000 men
versus 23 women in a sample of 1000 women were admitted because of heart disease. Do
these data present sufﬁcient evidence to indicate a higher rate of heart disease among men admitted to the hospital? Use a = 0,05. 19 Solution, We need to test the null hypothesis H0 3131 = P2
versus a one—tailed alternative Ha 3291 > P2"
The two separate sample proportions are A 3:1 52 A $2 23
z — z —— = 0052 d = — = —— = 01023
. M m 1000 an p2 n2 1000 and the pooled sample proportion is A 031 + x2 52 + 23
= = ————— = 00375“
p m + mg 1000 + 1000 So the test statistic is
A — A ., 2 — 1, 23
2=—————p1 p2 =—————————————005 00 =3‘41‘ me; + _1_) \/(0,0375)(019625) (331% + 13150) 712 Since a = 005, za 2 1645‘. Clearly, z = 3.41 > 11,645. Hence, H0 should be rejected,
* The critical value approach: For or = 0.05, za 2 1,645. We reject H0 if z > 1.645,
Since 341 > 11645, H0 is rejected. There is sufﬁcient evidence to indicate a higher rate of heart disease among men admitted to the hospital. * The pvalue approach: The p—value is
p—value = P{z > 3,41} = 1 — 09997 = 0,0003” f(z) Since jot—value: 00003 < 001, we would reject H0. There is sufﬁcient evidence to indicate a
higher rate of heart disease among men admitted to the hospital. HOMEWORK: 39191376 — 377
91.43, 9.45, 9,47, 9149, 9,51 20 ...
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 Spring '08
 Cheng
 Statistics

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